Solving Equations With Variables On Both Sides With Parentheses

Solving equations with variables on both sides, especially when parentheses are involved, is a fundamental skill in algebra. Mastering this technique opens doors to more complex mathematical concepts and real-world problem-solving. This comprehensive guide will walk you through the process step-by-step, providing clear explanations, examples, and strategies to conquer these types of equations.

Understanding the Basics

Before diving into the nitty-gritty, let's establish some foundational concepts. An equation is a mathematical statement that asserts the equality of two expressions. Solving an equation means finding the value(s) of the variable(s) that make the equation true. Equations with variables on both sides simply mean that the variable you're trying to solve for appears on both the left-hand side (LHS) and the right-hand side (RHS) of the equation. Parentheses, on the other hand, indicate a grouping of terms that need to be addressed before other operations can be performed.

The core principle we'll rely on is the concept of balancing the equation. Any operation performed on one side of the equation must be performed on the other side to maintain equality. This is often referred to as the golden rule of algebra. Common operations include:

• Addition: Adding the same value to both sides.
• Subtraction: Subtracting the same value from both sides.
• Multiplication: Multiplying both sides by the same non-zero value.
• Division: Dividing both sides by the same non-zero value.

The Step-by-Step Process: Solving Equations with Variables on Both Sides and Parentheses

Here's a breakdown of the systematic approach to tackling these equations:

Step 1: Distribute (Eliminate Parentheses)

This is often the first and most crucial step. The distributive property states that a(b + c) = ab + ac. In other words, you multiply the term outside the parentheses by each term inside the parentheses.

• Example: 2(x + 3) = 2x + 6

  Here, '2' is distributed to both 'x' and '3'.

• Important Considerations:

  • Be careful with negative signs! Remember that a negative sign in front of the parentheses applies to all terms inside. For example, -3(y - 2) = -3y + 6.
  • If there are multiple sets of parentheses, distribute each separately.

Step 2: Combine Like Terms on Each Side

After distributing, simplify each side of the equation by combining like terms. Like terms are terms that have the same variable raised to the same power (or are constants).

• Example: 3x + 2 - x + 5 = (3x - x) + (2 + 5) = 2x + 7

• Important Considerations:

  • Only combine terms on the same side of the equation. You cannot combine terms across the equals sign at this stage.
  • Pay attention to the signs of the terms when combining.

Step 3: Isolate the Variable Term on One Side

The goal now is to get all the variable terms on one side of the equation and all the constant terms on the other side. This is achieved by using addition or subtraction.

• Strategy: Look for the variable term with the smaller coefficient. Move this term to the other side of the equation. This often helps to avoid dealing with negative coefficients.

• Example: Suppose you have the equation 5x + 3 = 2x - 6. To move the '2x' term to the left side, subtract '2x' from both sides:

  5x + 3 - 2x = 2x - 6 - 2x

  This simplifies to: 3x + 3 = -6

• Important Considerations:

  • Remember to perform the same operation on both sides of the equation to maintain balance.
  • When moving a term from one side to the other, its sign changes (positive becomes negative, and vice versa).

Step 4: Isolate the Constant Term on the Other Side

Now that the variable terms are on one side, move all the constant terms to the other side using addition or subtraction.

• Example (Continuing from the previous example): We have 3x + 3 = -6. To move the '+3' to the right side, subtract '3' from both sides:

  3x + 3 - 3 = -6 - 3

  This simplifies to: 3x = -9

• Important Considerations:

  • Again, maintain balance by performing the same operation on both sides.
  • The sign of the constant term changes when moved.

Step 5: Solve for the Variable

The final step is to isolate the variable by dividing both sides of the equation by the coefficient of the variable.

• Example (Continuing from the previous example): We have 3x = -9. To solve for 'x', divide both sides by '3':

  3x / 3 = -9 / 3

  This simplifies to: x = -3

• Important Considerations:

  • Make sure you are dividing by the coefficient of the variable, not just any number.
  • Remember the rules of dividing positive and negative numbers.

Step 6: Check Your Solution (Optional but Recommended)

To ensure accuracy, substitute your solution back into the original equation and see if it holds true. If both sides of the equation are equal after the substitution, then your solution is correct.

• Example (Checking our solution): The original equation was 5x + 3 = 2x - 6, and we found x = -3. Substitute x = -3 into the original equation:

  5(-3) + 3 = 2(-3) - 6

  -15 + 3 = -6 - 6

  -12 = -12

  Since both sides are equal, our solution x = -3 is correct.

Examples with Detailed Explanations

Let's work through a few more examples to solidify the process:

Example 1: 4(2x - 1) + 5 = 3x + 10

1. Distribute:

   8x - 4 + 5 = 3x + 10

2. Combine Like Terms:

   8x + 1 = 3x + 10

3. Isolate Variable Term:

   8x - 3x + 1 = 3x - 3x + 10

   5x + 1 = 10

4. Isolate Constant Term:

   5x + 1 - 1 = 10 - 1

   5x = 9

5. Solve for Variable:

   5x / 5 = 9 / 5

   x = 9/5 or x = 1.8

6. Check (Optional):

   4(2(9/5) - 1) + 5 = 3(9/5) + 10

   4(18/5 - 5/5) + 5 = 27/5 + 50/5

   4(13/5) + 5 = 77/5

   52/5 + 25/5 = 77/5

   77/5 = 77/5 (Solution is correct)

Example 2: -2(3y + 4) - y = 5(y - 2) + 6

1. Distribute:

   -6y - 8 - y = 5y - 10 + 6

2. Combine Like Terms:

   -7y - 8 = 5y - 4

3. Isolate Variable Term:

   -7y - 5y - 8 = 5y - 5y - 4

   -12y - 8 = -4

4. Isolate Constant Term:

   -12y - 8 + 8 = -4 + 8

   -12y = 4

5. Solve for Variable:

   -12y / -12 = 4 / -12

   y = -1/3

6. Check (Optional):

   -2(3(-1/3) + 4) - (-1/3) = 5((-1/3) - 2) + 6

   -2(-1 + 4) + 1/3 = 5(-1/3 - 6/3) + 6

   -2(3) + 1/3 = 5(-7/3) + 6

   -6 + 1/3 = -35/3 + 18/3

   -18/3 + 1/3 = -17/3

   -17/3 = -17/3 (Solution is correct)

Example 3: 3(a - 2) + 4a = 2(5 - a) - 1

1. Distribute:

   3a - 6 + 4a = 10 - 2a - 1

2. Combine Like Terms:

   7a - 6 = 9 - 2a

3. Isolate Variable Term:

   7a + 2a - 6 = 9 - 2a + 2a

   9a - 6 = 9

4. Isolate Constant Term:

   9a - 6 + 6 = 9 + 6

   9a = 15

5. Solve for Variable:

   9a / 9 = 15 / 9

   a = 5/3

6. Check (Optional):

   3(5/3 - 2) + 4(5/3) = 2(5 - 5/3) - 1

   3(5/3 - 6/3) + 20/3 = 2(15/3 - 5/3) - 1

   3(-1/3) + 20/3 = 2(10/3) - 1

   -1 + 20/3 = 20/3 - 1

   -3/3 + 20/3 = 20/3 - 3/3

   17/3 = 17/3 (Solution is Correct)

Common Mistakes to Avoid

• Incorrect Distribution: Forgetting to distribute to all terms inside the parentheses, or mishandling negative signs during distribution.
• Combining Unlike Terms: Trying to add or subtract terms that don't have the same variable and exponent.
• Forgetting to Balance: Performing an operation on one side of the equation but not the other.
• Sign Errors: Making mistakes with positive and negative signs, especially when moving terms across the equals sign.
• Incorrect Order of Operations: Not following the correct order (PEMDAS/BODMAS) – Parentheses/Brackets, Exponents/Orders, Multiplication and Division (from left to right), Addition and Subtraction (from left to right).

Advanced Techniques and Considerations

• Fractions: If your equation contains fractions, you can often eliminate them by multiplying both sides of the equation by the least common multiple (LCM) of the denominators. This will result in an equation with integer coefficients.

• Decimals: Similarly, if you have decimals, you can multiply both sides of the equation by a power of 10 (10, 100, 1000, etc.) to eliminate the decimals.

• No Solution or Infinite Solutions: Some equations may have no solution (a contradiction) or infinite solutions (an identity).

  • No Solution: If, after simplifying, you arrive at a statement that is always false (e.g., 0 = 5), the equation has no solution.
  • Infinite Solutions: If, after simplifying, you arrive at a statement that is always true (e.g., 2 = 2), the equation has infinite solutions. This means that any value of the variable will satisfy the equation.

Conclusion

Solving equations with variables on both sides and parentheses requires a systematic approach, careful attention to detail, and a solid understanding of algebraic principles. By mastering the steps outlined in this guide, practicing regularly, and being mindful of common mistakes, you can confidently tackle these types of equations and build a strong foundation for further mathematical studies. Remember to always double-check your work, and don't be afraid to ask for help when needed. The more you practice, the easier it will become!